• No products in the cart.

Antipodal map and a fixed point theorem

Mathematicians love fixed point theorems. Brouwer fixed point theorem is one such result. It states that “for any continuous function f mapping a compact convex set into itself there is a point a such that f(a) = a.” We are interested in one such fixed point theorem.

Consider maps (continuous functions) from $$S^n \to S^n$$ (n-sphere to n-sphere). We will come up with a ‘test’ which will say, when do a function does not have a fixed point.

The antipodal map is defined as $$h(\bar{x}) = – \bar{x}$$. The bar above x says that may have more than one coordinate. For example a point on the 2-sphere has three coordinates (x, y, z). Hence h(x, y, z) = (-x, -y, – z).

In the above picture, we have drawn a 2-sphere. is any point on the sphere. is its antipodal point. B is found by changing the sign of each coordinate of A. Geometrically B is found by drawing a line from A, through the origin. Wherever that segment hits the sphere again, is its antipodal point.

It is easy to check that is a continuous map. Also, note that h does not have a fixed point. Since each point on the sphere is sent to some other point (the diametrically opposite point on the sphere), hence no point is left fixed by the map h.

There is a notion called homotopy of maps in advanced mathematics. The idea is this: suppose there are two maps and from X to Y. Can you continuously deform to ? If yes, then the two maps are said to be homotopic. In more technical language, and are homotopic if, there is a map $$F: X \times [0,1] \to Y$$ such that $$F(x, 0) = h(x), F(x, 1) = g(x)$$ and F is continuous.

(The word ‘map’ is used for ‘continuous function’).

Now we will show that any map which does not have a fixed point, is homotopic to the antipodal map.

Consider $$F(x, t) = \dfrac{(1-t) f(x) – tx} { || (1-t)f(x) – tx || }$$.

Here is a map from n-sphere to n-sphere. Here has n+1 coordinates.

Notice that F(x, 0) = f(x) (denominator is 1 as ||f(x)|| =1 , since f(x) is on the sphere).

Also F(x, 1) = – x, the antipodal map.

Clearly F is continous if the denominator is not zero. Let us find out, when the denominator is zero.

$$|| (1-t) f(x) – tx || = 0 \implies (1-t) f(x) = tx$$.

Taking the absolute value (norm) on both sides, we have $$|1-t| || f(x) || = |t| ||x||$$.

Since for and f(x)  are on the surface of the n-sphere, hence their length (norm or absolute value) is 1. Thus we have $$(1-t) = t$$.

This implies t = 1/2. But plugging in t = 1/2, in the equation (1-t) f(x) = tx, we have f(x) = x.  Hence , has a fixed point.

If did not have a fixed point then, the denominator will never be zero, making it homotopic to the antipodal map.

There is a beautiful visualization of this above fact. Suppose does not have a fixed point. Pick any point on the n-sphere. We want to send it to -a (the antipodal point). The shortest path is along the great circle, but there are infinitely many ways to reaching -a from a along a great circle as -a is exactly opposite to a.

Hence to decide on, which direction to go, we will take the help of f. Since does not have a fixed point, it will move to some a’. Once has moved a little in some direction (to a’), its path to -a becomes fixed (as it is nearer to -a along a certain great circle).

Hence any continuous function, from n-sphere to n-sphere, that does not have a fixed point is homotopic to (can be deformed continuously into) the antipodal map!

February 23, 2017